Bài 176 trang 67 SGK Toán 6 tập 2

Đề bài

Tính:

a) \(1{{13} \over {15}}.{\left( {0,5} \right)^2}.3 + \left( {{8 \over {15}} - 1{{19} \over {60}}} \right):1{{23} \over {24}}\)

b) \({{\left( {{{{{11}^2}} \over {200}} + 0,415} \right):0,01} \over {{1 \over {12}} - 37,25 + 3{1 \over 6}}}\)

Hướng dẫn giải

a) \(1{{13} \over {15}}.{\left( {0,5} \right)^2}.3 + \left( {{8 \over {15}} - 1{{19} \over {60}}} \right):1{{23} \over {24}}\)

\( = {{28} \over {15}}.{\left( {{1 \over 2}} \right)^2}.3 + \left( {{8 \over {15}} - {{79} \over {60}}} \right):{{47} \over {24}}\)

\( = {{28} \over {15}}.{1 \over 4}.3 + {{8.4 - 79} \over {60}}:{{47} \over {24}}\)

\( = {7 \over 5} + {{ - 47} \over {60}}.{{24} \over {47}}\)

\( = {7 \over 5} + {{ - 2} \over 5}\)

\( = {5 \over 5} = 1\)

b) \({{\left( {{{{{11}^2}} \over {200}} + 0,415} \right):0,01} \over {{1 \over {12}} - 37,25 + 3{1 \over 6}}}\)

\( = {{\left( {{{121} \over {200}} + {{415} \over {1000}}} \right):{1 \over {100}}} \over {{1 \over {12}} - {{149} \over 4} + {{19} \over 6}}}\)

\( = {{\left( {{{121} \over {200}} + {{83} \over {200}}} \right):{1 \over {100}}} \over {{{1 - 447 + 38} \over {12}}}}\)

\( = \left( {{{204} \over {200}}:{1 \over {100}}} \right):{{ - 408} \over {12}}\)

\( = {{102} \over {100}}.{{100} \over 1}.{{ - 12} \over {408}}\)

\( = {{ - 12} \over 4} =  - 3\)